$C H_{3} O C_{2} H_{5}$ and $(C H_{3})_{3} C - O C H_{3}$ are treated with hydroiodic acid. The fragments after reactions obtained are

C H_{3} I + H O C_{2} H_{5}; (C H_{3})_{3} C - I + H O C H_{3}

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C H_{3} O H + C_{2} H_{5} I; (C H_{3})_{3} C - I + H O C H_{3}

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C H_{3} O H + C_{2} H_{5} I; (C H_{3})_{3} C - O H + C H_{3} I

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C H_{3} I + C_{2} H_{5} O H; (C H_{3})_{3} C - O H + C H_{3} I

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Solution

The correct option is A $C H_{3} I + H O C_{2} H_{5}; (C H_{3})_{3} C - I + H O C H_{3}$
When mixed ethers are used, the alkyl iodide produced depends on the nature of alkyl groups. If one group is $M e$ and the other a pri- or sec- alkyl group, then methyl iodide is produced. Here reaction occurs via $S_{N} 2$ mechanism and because of the steric effect of the larger group, $I^{-}$ attacks the smaller $(M e)$ group.
$C H_{3} O C_{2} H_{5} + H I \to C H_{3} I + C_{2} H_{5} O H$

When the substrate is a methyl tert-alkyl ethers, the products are tert-RI and $M e O H$ . Here reaction occurs by $S_{N} 1$ mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is
$3^{o} > 2^{o} > 1^{o} > + C H_{3}$

$∴$
Alkyl halide is always derived from tert-alkyl group.

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